JEE Main 2022 (Online) 24th June Morning ShiftQuestion 1
Let $S$ be the region bounded by the curves $y=x^{3}$ and $y^{2}=x$. The curve $y=2|x|$ divides $S$ into two regions of areas $R_1$ and $R_2$. If $\max\{R_1,R_2\}=R_2$, then $\frac{R_2}{R_1}$ is equal to $\_\_\_\_\_\_$ .
Application of Integrals
Answer$19$
Solution
$S$ lies in the first quadrant between $y=x^3$ and $y=\sqrt x$, $0\le x\le1$: area $=\int_0^1(\sqrt x-x^3)dx=\frac23-\frac14=\frac5{12}$. $y=2x$ meets $y=\sqrt x$ at $4x^2=x\Rightarrow x=\frac14$ (and meets $y=x^3$ only at $0$ inside $S$). The part of $S$ above the line is $\int_0^{1/4}(\sqrt x-2x)dx=\frac23\cdot\frac18-\frac1{16}=\frac1{48}$. So $R_1=\frac1{48}$, $R_2=\frac5{12}-\frac1{48}=\frac{19}{48}$. $\frac{R_2}{R_1}=19$.